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Difference between revisions of "Mock AIME I 2012 Problems/Problem 10"

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(Solution 2)
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== Problem 10 ==
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== Problem ==
 
Consider the function <math>f(n,x) = \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{(n-1)x} + \sin{nx}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{(n-1)x} + \cos{nx}}</math>. Find the sum of all <math>x</math> for which <math>f(23,x)=f(33,x)</math>, where <math>x</math> is measured in degrees and <math>100<x<200</math>.
 
Consider the function <math>f(n,x) = \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{(n-1)x} + \sin{nx}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{(n-1)x} + \cos{nx}}</math>. Find the sum of all <math>x</math> for which <math>f(23,x)=f(33,x)</math>, where <math>x</math> is measured in degrees and <math>100<x<200</math>.
  
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== Solution 1 ==
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== Solution 1 (Sum-to-Product) ==
 
Recalling the trigonometric [[Trigonometric identities#Sum-to-product identities|sum-to-product identities]], we can rearrange terms and evaluate <math>f(23,x)</math> as follows:
 
Recalling the trigonometric [[Trigonometric identities#Sum-to-product identities|sum-to-product identities]], we can rearrange terms and evaluate <math>f(23,x)</math> as follows:
 
\begin{align*}
 
\begin{align*}
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Now, we desire to find all <math>x\in(100^{\circ},200^{\circ})</math> that satisfy the following equation: <cmath>f(22,x)=f(33,x)\iff\tan{12x}=\tan{17x}.</cmath> Because <math>\tan x</math> has a period of <math>180^{\circ}</math> and it only reaches any given value once per period (by virtue of being monotone increasing between its asymptotes), we know that <math>5x</math> must then be some integer multiple of tangent's period, <math>180^{\circ}</math>. Thus, <math>x</math> must be a multiple of <math>\tfrac{180^{\circ}}5=36^{\circ}</math>, and so the possible values of <math>x</math> between <math>100^{\circ}</math> and <math>200^{\circ}</math> are <math>108^{\circ}</math>, <math>144^{\circ}</math>, and <math>180^{\circ}</math>.
 
Now, we desire to find all <math>x\in(100^{\circ},200^{\circ})</math> that satisfy the following equation: <cmath>f(22,x)=f(33,x)\iff\tan{12x}=\tan{17x}.</cmath> Because <math>\tan x</math> has a period of <math>180^{\circ}</math> and it only reaches any given value once per period (by virtue of being monotone increasing between its asymptotes), we know that <math>5x</math> must then be some integer multiple of tangent's period, <math>180^{\circ}</math>. Thus, <math>x</math> must be a multiple of <math>\tfrac{180^{\circ}}5=36^{\circ}</math>, and so the possible values of <math>x</math> between <math>100^{\circ}</math> and <math>200^{\circ}</math> are <math>108^{\circ}</math>, <math>144^{\circ}</math>, and <math>180^{\circ}</math>.
  
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Now, we can add these three values to compute our final answer:
+
Now, after checking for extraneous solutions (of which there are none), we can add these three values to compute our final answer:
 +
<cmath>108+144+180=\boxed{432}.</cmath>
 +
 
 +
== Solution 2 (Integral Calculus, non-rigorous) ==
 +
Recall that the average value of some function <math>g(x)</math> over the interval <math>[a,b]</math> is given by <cmath>\frac1{b-a}\int_a^b g(x)dx.</cmath> Then, we can approximate (hopefully well enough) the above expression for <math>f(n,x)</math> by multiplying <math>n</math> by the average of the function <math>\sin(kx)</math> over the interval <math>[0,n]</math>:
 +
\begin{align*}
 +
f(n,x) &= \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{(n-1)x} + \sin{nx}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{(n-1)x} + \cos{nx}} \\
 +
&\approx \dfrac{n(\frac1n\int_0^n\sin(kx)dk)}{n(\frac1n\int_0^n\cos(kx)dk)} \\
 +
&= \dfrac{-(\frac1x\cos(kx))|^n_0}{(\frac1x\sin(kx))|^n_0} \\
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&= \dfrac{1-\cos(nx)}{\sin(nx)}
 +
\end{align*}
 +
 
 +
Now, we desire to find <math>x\in(100^{\circ},200^{\circ})</math> that satisfy the equation: <cmath>f(23,x)=f(33,x)\iff\frac{1-\cos(23x)}{\sin(23x)}=\frac{1-\cos(33x)}{\sin(33x)}.</cmath> Recalling the [[Trigonometric identities#Half-angle identities|tangent half-angle identities]], we can solve as follows:
 +
\begin{align*}
 +
\frac{1-\cos(23x)}{\sin(23x)} &= \frac{1-\cos(33x)}{\sin(33x)} \\
 +
\tan\left(\frac{23}2x\right) &= \tan\left(\frac{33}2x\right) = \tan\left(\frac{23}2x+5x\right).
 +
\end{align*}
 +
 
 +
Now, because <math>\tan x</math> has a period of <math>180^{\circ}</math> and it only reaches any given value once per period (by virtue of being monotone increasing between its asymptotes), we know that <math>5x</math> must then be some integer multiple of tangent's period, <math>180^{\circ}</math>. Thus, <math>x</math> must be a multiple of <math>\tfrac{180^{\circ}}5=36^{\circ}</math>, and so the possible values of <math>x</math> between <math>100^{\circ}</math> and <math>200^{\circ}</math> are <math>108^{\circ}</math>, <math>144^{\circ}</math>, and <math>180^{\circ}</math>.
 +
 
 +
Now, after checking for extraneous solutions (of which there are none), we can add these three values to compute our final answer:
 
<cmath>108+144+180=\boxed{432}.</cmath>
 
<cmath>108+144+180=\boxed{432}.</cmath>

Revision as of 19:35, 10 March 2025

Problem

Consider the function $f(n,x) = \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{(n-1)x} + \sin{nx}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{(n-1)x} + \cos{nx}}$. Find the sum of all $x$ for which $f(23,x)=f(33,x)$, where $x$ is measured in degrees and $100<x<200$.

Solution 1 (Sum-to-Product)

Recalling the trigonometric sum-to-product identities, we can rearrange terms and evaluate $f(23,x)$ as follows: \begin{align*} f(23, x) &= \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{22x} + \sin{23x}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{22x} + \cos{23x}} \\ &= \dfrac{(\sin{x} + \sin{23x}) + (\sin{2x} + \sin{22x}) + \cdots + (\sin{11x} + \sin{13x}) + \sin{12x}}{(\cos{x} + \cos{23x}) + (\cos{2x} + \cos{22x}) + \cdots + (\cos{11x} + \cos{13x}) + \cos{12x}} \\ &= \dfrac{2\sin{12x}\cos{11x}+2\sin{12x}\cos{10x}+\cdots+2\sin{12x}\cos{x}+\sin{12x}}{2\cos{12x}\cos{11x}+2\cos{12x}\cos{10x}+\cdots+2\cos{12x}\cos{x}+\cos{12x}} \\ &= \dfrac{2\sin{12x}(\cos{11x}+\cos{10x}+\cos{9x}+\cdots+\cos{x}+1)}{2\cos{12x}(\cos{11x}+\cos{10x}+\cdots+\cos{x}+1)} \\ &= \dfrac{\sin{12x}}{\cos{12x}} \\ &= \tan{12x} \end{align*}

Likewise, we can show that \[f(33,x)=\dfrac{\sin{17x}}{\cos{17x}}=\tan{17x}.\]

Now, we desire to find all $x\in(100^{\circ},200^{\circ})$ that satisfy the following equation: \[f(22,x)=f(33,x)\iff\tan{12x}=\tan{17x}.\] Because $\tan x$ has a period of $180^{\circ}$ and it only reaches any given value once per period (by virtue of being monotone increasing between its asymptotes), we know that $5x$ must then be some integer multiple of tangent's period, $180^{\circ}$. Thus, $x$ must be a multiple of $\tfrac{180^{\circ}}5=36^{\circ}$, and so the possible values of $x$ between $100^{\circ}$ and $200^{\circ}$ are $108^{\circ}$, $144^{\circ}$, and $180^{\circ}$.

Now, after checking for extraneous solutions (of which there are none), we can add these three values to compute our final answer: \[108+144+180=\boxed{432}.\]

Solution 2 (Integral Calculus, non-rigorous)

Recall that the average value of some function $g(x)$ over the interval $[a,b]$ is given by \[\frac1{b-a}\int_a^b g(x)dx.\] Then, we can approximate (hopefully well enough) the above expression for $f(n,x)$ by multiplying $n$ by the average of the function $\sin(kx)$ over the interval $[0,n]$: \begin{align*} f(n,x) &= \dfrac{\sin{x} + \sin{2x} + \sin{3x} + \cdots + \sin{(n-1)x} + \sin{nx}}{\cos{x} + \cos{2x} + \cos{3x} + \cdots + \cos{(n-1)x} + \cos{nx}} \\ &\approx \dfrac{n(\frac1n\int_0^n\sin(kx)dk)}{n(\frac1n\int_0^n\cos(kx)dk)} \\ &= \dfrac{-(\frac1x\cos(kx))|^n_0}{(\frac1x\sin(kx))|^n_0} \\ &= \dfrac{1-\cos(nx)}{\sin(nx)} \end{align*}

Now, we desire to find $x\in(100^{\circ},200^{\circ})$ that satisfy the equation: \[f(23,x)=f(33,x)\iff\frac{1-\cos(23x)}{\sin(23x)}=\frac{1-\cos(33x)}{\sin(33x)}.\] Recalling the tangent half-angle identities, we can solve as follows: \begin{align*} \frac{1-\cos(23x)}{\sin(23x)} &= \frac{1-\cos(33x)}{\sin(33x)} \\ \tan\left(\frac{23}2x\right) &= \tan\left(\frac{33}2x\right) = \tan\left(\frac{23}2x+5x\right). \end{align*}

Now, because $\tan x$ has a period of $180^{\circ}$ and it only reaches any given value once per period (by virtue of being monotone increasing between its asymptotes), we know that $5x$ must then be some integer multiple of tangent's period, $180^{\circ}$. Thus, $x$ must be a multiple of $\tfrac{180^{\circ}}5=36^{\circ}$, and so the possible values of $x$ between $100^{\circ}$ and $200^{\circ}$ are $108^{\circ}$, $144^{\circ}$, and $180^{\circ}$.

Now, after checking for extraneous solutions (of which there are none), we can add these three values to compute our final answer: \[108+144+180=\boxed{432}.\]

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